The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?

We start by counting the ways to choose three elements:

5

·

4

·

3 = 60

The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?

We start by counting the ways to choose three elements:

5

·

4

·

3 = 60

The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?We start by counting the ways to choose three elements:

5

·

4

·

3 = 60

The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?We start by counting the ways to choose three elements:

5 ·

4

·

3 = 60

The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?We start by counting the ways to choose three elements:

5 · 4 ·

3 = 60

The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?We start by counting the ways to choose three elements:

5 · 4 · 3

= 60

The ProblemHow many ways are there to pick a committee of three frompersons A, B, C, D, and E?

A Simpler ProblemHow many ways to line up three of those people in order?We start by counting the ways to choose three elements:

5 · 4 · 3 = 60

Sixty committees?

ABCABDABEACBACDACEADBADCADEAEBAECAED

BACBADBAEBCABCDBCEBDABDCBDEBEABECBED

CABCADCAECBACBDCBECDACDBCDECEACEBCED

DABDACDAEDBADBCDBEDCADCBDCEDEADEBDEC

EABEACEADEBAEBCEBDECAECBECDEDAEDBEDC

Sixty committees?

ABCABDABEACBACDACEADBADCADEAEBAECAED

BACBADBAEBCABCDBCEBDABDCBDEBEABECBED

CABCADCAECBACBDCBECDACDBCDECEACEBCED

DABDACDAEDBADBCDBEDCADCBDCEDEADEBDEC

EABEACEADEBAEBCEBDECAECBECDEDAEDBEDC

Sixty committees?

ABCABDABEACBACDACEADBADCADEAEBAECAED

BACBADBAEBCABCDBCEBDABDCBDEBEABECBED

CABCADCAECBACBDCBECDACDBCDECEACEBCED

DABDACDAEDBADBCDBEDCADCBDCEDEADEBDEC

EABEACEADEBAEBCEBDECAECBECDEDAEDBEDC

No! These six are the same!

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2 · 1 = 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2 · 1 = 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2 · 1 = 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2 · 1 = 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3

· 2 · 1 = 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2

· 1 = 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2 · 1

= 3! = 6

Why 6?The single committee {A,B,C} shows up once for eachway it can be listed:ABC ACB BAC BCA CAB CBA

How many ways can those three people be listed?

3 · 2 · 1 = 3! = 6

So how many?

ABCABDABEACBACDACEADBADCADEAEBAECAED

BACBADBAEBCABCDBCEBDABDCBDEBEABECBED

CABCADCAECBACBDCBECDACDBCDECEACEBCED

DABDACDAEDBADBCDBEDCADCBDCEDEADEBDEC

EABEACEADEBAEBCEBDECAECBECDEDAEDBEDC

So how many?

ABCABDABEACBACDACEADBADCADEAEBAECAED

BACBADBAEBCABCDBCEBDABDCBDEBEABECBED

CABCADCAECBACBDCBECDACDBCDECEACEBCED

DABDACDAEDBADBCDBEDCADCBDCEDEADEBDEC

EABEACEADEBAEBCEBDECAECBECDEDAEDBEDC

So how many?

ABCABDABEACBACDACEADBADCADEAEBAECAED

BACBADBAEBCABCDBCEBDABDCBDEBEABECBED

CABCADCAECBACBDCBECDACDBCDECEACEBCED

DABDACDAEDBADBCDBEDCADCBDCEDEADEBDEC

EABEACEADEBAEBCEBDECAECBECDEDAEDBEDC

So how many?

Ten!

ABC

ABD

ABE

ACB

ACD

ACE

ADB

ADC

ADE

AEB

AEC

AED

BAC

BAD

BAE

BCA

BCD

BCE

BDA

BDC

BDE

BEA

BEC

BED

CAB

CAD

CAE

CBA

CBD

CBE

CDA

CDB

CDE

CEA

CEB

CED

DAB

DAC

DAE

DBA

DBC

DBE

DCA

DCB

DCE

DEA

DEB

DEC

EAB

EAC

EAD

EBA

EBC

EBD

ECA

ECB

ECD

EDA

EDB

EDC

{A,B,C}{A,B,D}{A,B,E}{A,C,D}{A,C,E}{A,D,E}{B,C,D}{B,C,E}{B,D,E}{C,D,E}

Ten!

ABC

ABD

ABE

ACB

ACD

ACE

ADB

ADC

ADE

AEB

AEC

AED

BAC

BAD

BAE

BCA

BCD

BCE

BDA

BDC

BDE

BEA

BEC

BED

CAB

CAD

CAE

CBA

CBD

CBE

CDA

CDB

CDE

CEA

CEB

CED

DAB

DAC

DAE

DBA

DBC

DBE

DCA

DCB

DCE

DEA

DEB

DEC

EAB

EAC

EAD

EBA

EBC

EBD

ECA

ECB

ECD

EDA

EDB

EDC

{A,B,C}{A,B,D}{A,B,E}{A,C,D}{A,C,E}{A,D,E}{B,C,D}{B,C,E}{B,D,E}{C,D,E}

603!

= 10

What Happened?P(5,3) = 5 · 4 · 3 = 60 lists of three peopleEach committee of three shows up in the listP(3,3) = 3! = 6 timesTherefore the number of committees is

P(5,3)P(3,3)

=606

= 10.

What Happened?P(5,3) = 5 · 4 · 3 = 60 lists of three peopleEach committee of three shows up in the listP(3,3) = 3! = 6 timesTherefore the number of committees is

P(5,3)P(3,3)

=606

= 10.

What Happened?P(5,3) = 5 · 4 · 3 = 60 lists of three peopleEach committee of three shows up in the listP(3,3) = 3! = 6 timesTherefore the number of committees is

P(5,3)P(3,3)

=606

= 10.

ConclusionsIn general, suppose you have n people and you want tochoose r of them for a committee.Then there are P(n, r) ways to choose them in order.Each committee shows up in that list P(r , r) = r ! times.Therefore the number of committees is

C(n, r) =

P(n, r)r !

This is called the number of combinations of n objectstaken r at a time.It is written as C(n, r) and is pronounced “n choose r .”

ConclusionsIn general, suppose you have n people and you want tochoose r of them for a committee.Then there are P(n, r) ways to choose them in order.Each committee shows up in that list P(r , r) = r ! times.Therefore the number of committees is

C(n, r) =

P(n, r)r !

This is called the number of combinations of n objectstaken r at a time.It is written as C(n, r) and is pronounced “n choose r .”

ConclusionsIn general, suppose you have n people and you want tochoose r of them for a committee.Then there are P(n, r) ways to choose them in order.Each committee shows up in that list P(r , r) = r ! times.Therefore the number of committees is

C(n, r) =

P(n, r)r !

This is called the number of combinations of n objectstaken r at a time.It is written as C(n, r) and is pronounced “n choose r .”

ConclusionsIn general, suppose you have n people and you want tochoose r of them for a committee.Then there are P(n, r) ways to choose them in order.Each committee shows up in that list P(r , r) = r ! times.Therefore the number of committees is

C(n, r) =

P(n, r)r !

This is called the number of combinations of n objectstaken r at a time.It is written as C(n, r) and is pronounced “n choose r .”

ConclusionsIn general, suppose you have n people and you want tochoose r of them for a committee.Then there are P(n, r) ways to choose them in order.Each committee shows up in that list P(r , r) = r ! times.Therefore the number of committees is

C(n, r) =

P(n, r)r !

This is called the number of combinations of n objectstaken r at a time.It is written as C(n, r) and is pronounced “n choose r .”

ConclusionsIn general, suppose you have n people and you want tochoose r of them for a committee.Then there are P(n, r) ways to choose them in order.Each committee shows up in that list P(r , r) = r ! times.Therefore the number of committees is

C(n, r) =P(n, r)

r !

This is called the number of combinations of n objectstaken r at a time.It is written as C(n, r) and is pronounced “n choose r .”

Permutations vs. Combinations

Important FactThe difference between permutations and combinations is theidea of order.

Key QuestionsWhen selecting objects, ask yourself:

Would it count as a different result if I rearranged theobjects I chose?Am I selecting these objects to play different roles in theproblem?

If the answer to either question is yes, use Permutations; if theanswer to both questions is no, use Combinations.

Permutations vs. Combinations

Important FactThe difference between permutations and combinations is theidea of order.

Key QuestionsWhen selecting objects, ask yourself:

Would it count as a different result if I rearranged theobjects I chose?Am I selecting these objects to play different roles in theproblem?

If the answer to either question is yes, use Permutations; if theanswer to both questions is no, use Combinations.

Practice Problems(cf. 5.5 #39–43)

1 Suppose that you have 35 songs on your MP3 player. Howmany ways can you make a playlist of 5 songs?

2 Of the 20 applicants for a job, 4 will be selected forintensive interviews. In how many ways can the selectionbe made?

3 In a batch of 100 USB drives, 7 are defective. A sample ofthree drives is to be selected from the batch. How manysamples are possible? How many of the samples consistof all defective drives?

4 A student must choose five courses out of seven that hewould like to take. How many possibilities are there?

5 How many different three-letter “words” are there having norepetition of letters?

A More Advanced Problem

6 A club has six people: Alice, Bob, Charlie, Danielle, Edgar,and Frederika. Alice and Danielle are twins, and so areBob and Charlie. How many ways are there to line up thesix people in a row for a picture, if we insist that each pairof twins is standing together?